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The Khovanov–Seidel bimodule maps β_i and γ_i

Definition

Fix $m \geq 1$ , let $A_{m}$ be the Khovanov–Seidel type A algebra with its internal grading and its left-to-right path multiplication of Khovanov–Seidel type A algebra and Integral path ring of a finite quiver, and for $1 \leq i \leq m$ let $P_{i} = A_{m} e_{i} (paths ending at i),_{i} P = e_{i} A_{m} (paths beginning at i), U_{i} = P_{i} \otimes_{Z}_{i} P$ be the graded $(A_{m}, A_{m})$ -bimodule of The two-sided projective bimodules U_i and their tensor functors, with left action $a \cdot (x \otimes y) = (a x) \otimes y$ and right action $(x \otimes y) \cdot a = x \otimes (y a)$ .

The multiplication map. Let $β_{i} : U_{i} ⟶ A_{m}, β_{i} (x \otimes y) : = x y,$ be the $Z$ -bilinear extension of the product in $A_{m}$ . It is the degree-zero $(A_{m}, A_{m})$ -bimodule map $U_{i} \to A_{m}$ with $β_{i} (e_{i} \otimes e_{i}) = e_{i}$ , and it is uniquely determined by that value, as proved below. This is equation (2.6) of the source.

The map $γ_{i}$ . Let $w_{i} : = (i - 1 ∣ i) \otimes (i ∣ i - 1) + (i + 1 ∣ i) \otimes (i ∣ i + 1) + (i) \otimes (i ∣ i - 1 ∣ i) + (i ∣ i - 1 ∣ i) \otimes (i) \in U_{i}$ be the sum of the four displayed elementary tensors of $U_{i}$ , with the second summand $(i + 1 ∣ i) \otimes (i ∣ i + 1)$ omitted when $i = m$ , so that $w_{m} = (m - 1 ∣ m) \otimes (m ∣ m - 1) + (m) \otimes (m ∣ m - 1 ∣ m) + (m ∣ m - 1 ∣ m) \otimes (m);$ the omission is forced, because the arrow $(i + 1 ∣ i)$ exists only for $i \leq m - 1$ . Let $γ_{i} : A_{m} ⟶ U_{i} {- 1}, γ_{i} (a) : = a \cdot w_{i},$ the left action of $A_{m}$ on $U_{i}$ followed by the identification with the internal shift $U_{i} {- 1}$ of Finite graded A_m-modules, internal shifts and the vertex projectives; this is equation (2.7) of the source.

Claims proved below. The element $w_{i}$ lies in $U_{i}$ and is homogeneous of internal degree $1$ , so that $γ_{i}$ takes values in $U_{i} {- 1}$ and $γ_{i} (1) = w_{i}$ has degree $0$ there; $w_{i}$ is central, $a \cdot w_{i} = w_{i} \cdot a$ for every $a \in A_{m}$ , so $γ_{i}$ is a map of $(A_{m}, A_{m})$ -bimodules and not only left $A_{m}$ -linear; $β_{i}$ is a well-defined degree-zero $(A_{m}, A_{m})$ -bimodule map; and consequently $β_{i}$ and $γ_{i}$ induce natural transformations $U_{i} (-) \to I d$ and $I d \to U_{i} {- 1} (-)$ on the category $A_{m} -mod$ .

Convention. The shift is the internal one, $(U_{i} {- 1})_{d} = (U_{i})_{d + 1}$ , and never the homological shift $[1]$ ; both $β_{i}$ and $γ_{i}$ are degree-zero maps of graded bimodules, the shift in the target of $γ_{i}$ absorbing the degree one of $w_{i}$ . Indices run over $1 \leq i \leq m$ as in the source, and the term omitted at $i = m$ is the only one that involves a non-existent arrow.

Facts & Assumptions

Given: An integer $m \geq 1$ , the algebra $A_{m}$ with vertex idempotents $e_{j}$ , arrows $(j ∣ j \pm 1)$ , returns $(j ∣ j - 1 ∣ j)$ , its internal grading, and the bimodules $P_{i} = A_{m} e_{i}$ , $_{i} P = e_{i} A_{m}$ , $U_{i} = P_{i} \otimes_{Z}_{i} P$ for an index $1 \leq i \leq m$ .

[F1]

The product of two paths in $Z Γ_{m}$ is their left-to-right concatenation when they compose and $0$ otherwise; the unit is $\sum_{j = 0}^{m} (j)$ ; $(j) p = p$ exactly for paths $p$ beginning at $j$ and $p (j) = p$ exactly for paths $p$ ending at $j$ ; consequently $P_{j} = A_{m} e_{j}$ is the subgroup spanned by the paths ending at $j$ and $_{j} P = e_{j} A_{m}$ the subgroup spanned by the paths beginning at $j$ (Integral path ring of a finite quiver, Finite graded A_m-modules, internal shifts and the vertex projectives).

[L2]

$A_{m}$ has the $Z$ -basis of $4 m + 1$ classes given by the vertices, the $2 m$ arrows and the returns $(1 ∣ 0 ∣ 1), \dots, (m ∣ m - 1 ∣ m)$ ; every path of length at least three has class $0$ ; the monotone length-two paths $(j ∣ j + 1 ∣ j + 2)$ and $(j + 2 ∣ j + 1 ∣ j)$ have class $0$ for $0 \leq j \leq m - 2$ ; the return $(0 ∣ 1 ∣ 0)$ has class $0$ ; and at an interior vertex $0 < j < m$ the two returns agree, $(j ∣ j + 1 ∣ j) = (j ∣ j - 1 ∣ j)$ (The 4m+1 path basis).

[F3]

The internal degree is additive over concatenation, with $\deg (j) = \deg (j ∣ j + 1) = 0$ and $\deg (j + 1 ∣ j) = 1$ for all $0 \leq j \leq m - 1$ ; in particular $\deg (i ∣ i - 1 ∣ i) = 1$ for $1 \leq i \leq m$ (Khovanov–Seidel type A algebra).

[L4]

For graded modules the balanced tensor carries the total grading in which a homogeneous elementary tensor has degree the sum of the degrees of its factors; $U_{i} = P_{i} \otimes_{Z}_{i} P$ is a graded $(A_{m}, A_{m})$ -bimodule with the actions displayed in the definition, and every elementary tensor is a finite sum of such homogeneous terms (Graded balanced tensor product and homogeneous Hom, The two-sided projective bimodules U_i and their tensor functors).

[L5]

For a graded ring $R$ and a graded left $R$ -module $N$ the unit map $R \otimes_{R} N \to N$ , $r \otimes n \mapsto r n$ , is a natural degree-zero isomorphism of graded left $R$ -modules, and the balanced tensor is functorial in each variable (Graded associativity, units, and internal-shift tensor isomorphisms).

Proof

technique · direct

1.1

The displayed tensors lie in $U_{i}$ . By [F1] the module $P_{i}$ is spanned by the paths ending at $i$ and $_{i} P$ by the paths beginning at $i$ ; each first factor occurring in $w_{i}$ , namely $(i - 1 ∣ i)$ , $(i + 1 ∣ i)$ , $(i)$ and $(i ∣ i - 1 ∣ i)$ , ends at $i$ and so lies in $P_{i}$ , and each second factor, namely $(i ∣ i - 1)$ , $(i ∣ i + 1)$ , $(i ∣ i - 1 ∣ i)$ and $(i)$ , begins at $i$ and so lies in $_{i} P$ ; hence every displayed elementary tensor is an element of $U_{i} = P_{i} \otimes_{Z}_{i} P$ and $w_{i}$ is a well-defined element of $U_{i}$ . The summand $(i + 1 ∣ i) \otimes (i ∣ i + 1)$ involves the arrow $(i + 1 ∣ i)$ , which for $i = m$ is not an arrow of the quiver, so the omission at $i = m$ is forced and not a convention.

F1L4

1.2

Degree of $w_{i}$ . By [F3] the degrees of the four first factors are $0, 1, 0, 1$ and those of the four second factors are $1, 0, 1, 0$ , so by the additivity of the degree over concatenation and the total grading of [L4] the four summands of $w_{i}$ have degrees $0 + 1$ , $1 + 0$ , $0 + 1$ and $1 + 0$ , all equal to $1$ ; hence $w_{i}$ is homogeneous of degree $1$ in $U_{i}$ , and in the shifted module $U_{i} {- 1}$ , where degrees are lowered by one, the element $w_{i} = γ_{i} (1)$ has degree $0$ , matching the degree $0$ of $1 \in A_{m}$ .

F3L4

1.3

$β_{i}$ is a degree-zero bimodule map. Multiplication $A_{m} \times A_{m} \to A_{m}$ is $Z$ -bilinear and hence induces a well-defined $Z$ -linear map on the tensor product, and it is degree zero because $\deg (x y) = \deg (x) + \deg (y)$ by [F3] is exactly the degree of the elementary tensor $x \otimes y$ in the total grading of [L4]; moreover for $a, b \in A_{m}$ and an elementary tensor $x \otimes y \in U_{i}$ one has $β_{i} (a \cdot (x \otimes y) \cdot b) = β_{i} ((a x) \otimes (y b)) = (a x) (y b) = a (x y) b = a β_{i} (x \otimes y) b$ by associativity of $A_{m}$ , so $β_{i}$ is left and right $A_{m}$ -linear.

F1F3L4

2.1

$β_{i}$ is the unique bimodule map with $β_{i} (e_{i} \otimes e_{i}) = e_{i}$ . By [F1] every $x \in P_{i}$ satisfies $x e_{i} = x$ and every $y \in_{i} P$ satisfies $e_{i} y = y$ , so $x \otimes y = x \cdot (e_{i} \otimes e_{i}) \cdot y$ for each elementary tensor; a bimodule map $β : U_{i} \to A_{m}$ with $β (e_{i} \otimes e_{i}) = e_{i}$ therefore satisfies $β (x \otimes y) = x e_{i} y = x y = β_{i} (x \otimes y)$ on elementary tensors, hence on all of $U_{i}$ by additivity, so $β = β_{i}$ and in particular $β_{i} (e_{i} \otimes e_{i}) = e_{i} \neq 0$ .

step 1.3F1

2.2

Weight form of the centrality identity. Write $w_{i} = \sum_{k} x_{k} \otimes y_{k}$ for the at most four displayed elementary tensors, so each $x_{k} \in P_{i}$ begins at a vertex $s_{k} \in {i - 1, i, i + 1}$ and each $y_{k} \in_{i} P$ ends at that same vertex $s_{k}$ , the value $i + 1$ not occurring when $i = m$ ; for a path $p \in A_{m}$ the left action gives $p \cdot (x_{k} \otimes y_{k}) = (p x_{k}) \otimes y_{k}$ , which is $0$ unless $t (p) = s_{k}$ , and the right action gives $(x_{k} \otimes y_{k}) \cdot p = x_{k} \otimes (y_{k} p)$ , which is $0$ unless $s (p) = s_{k}$ . Since $s (p) \neq t (p)$ for an arrow, the index $k$ contributes to $p \cdot w_{i}$ or to $w_{i} \cdot p$ but never to both, so $p \cdot w_{i} = \sum_{k : t (p) = s_{k}} (p x_{k}) \otimes y_{k}$ and $w_{i} \cdot p = \sum_{k : s (p) = s_{k}} x_{k} \otimes (y_{k} p)$ ; centrality is therefore the finite list of checks on vertex idempotents and arrows carried out in steps 3.1, 3.2, 3.3, 3.4, 3.5, 3.6 and 3.7 below, and by $Z$ -linearity in $p$ it suffices to run them on the path generators of $A_{m}$ .

F1step 1.1

3.1

Vertex idempotents. For every vertex idempotent $e_{j}$ and every $k$ one has $e_{j} x_{k} = x_{k}$ and $y_{k} e_{j} = y_{k}$ when $s_{k} = j$ , and $e_{j} x_{k} = 0 = y_{k} e_{j}$ otherwise, by [F1] and the description of $s_{k}$ in step 2.2; summing over $k$ gives $e_{j} \cdot w_{i} = \sum_{k : s_{k} = j} x_{k} \otimes y_{k} = w_{i} \cdot e_{j}$ , so $w_{i}$ centralizes every vertex idempotent and hence every $Z$ -linear combination of them.

step 2.2F1

3.2

The arrow $(i - 1 ∣ i)$ . Here $t (p) = i$ and $s (p) = i - 1$ , and the only terms with $s_{k} = i$ are $(i) \otimes (i ∣ i - 1 ∣ i)$ and $(i ∣ i - 1 ∣ i) \otimes (i)$ , so $p \cdot w_{i} = ((i - 1 ∣ i) (i)) \otimes (i ∣ i - 1 ∣ i) + ((i - 1 ∣ i) (i ∣ i - 1 ∣ i)) \otimes (i) = (i - 1 ∣ i) \otimes (i ∣ i - 1 ∣ i)$ , the second summand being $0$ because $(i - 1 ∣ i ∣ i - 1 ∣ i)$ is a path of length three; and $w_{i} \cdot p = ((i - 1 ∣ i) \otimes (i ∣ i - 1)) \cdot p = (i - 1 ∣ i) \otimes ((i ∣ i - 1) (i - 1 ∣ i)) = (i - 1 ∣ i) \otimes (i ∣ i - 1 ∣ i)$ . The two sides agree, and this arrow exists for every $1 \leq i \leq m$ .

step 2.2L2

3.3

The arrow $(i ∣ i - 1)$ . Here $t (p) = i - 1$ and $s (p) = i$ , so $p \cdot w_{i} = ((i ∣ i - 1) (i - 1 ∣ i)) \otimes (i ∣ i - 1) = (i ∣ i - 1 ∣ i) \otimes (i ∣ i - 1)$ , while $w_{i} \cdot p = ((i) \otimes ((i ∣ i - 1 ∣ i) (i ∣ i - 1))) + ((i ∣ i - 1 ∣ i) \otimes ((i) (i ∣ i - 1))) = (i ∣ i - 1 ∣ i) \otimes (i ∣ i - 1)$ , the first summand being $0$ because the path $(i ∣ i - 1 ∣ i ∣ i - 1)$ has length three and the second being the displayed term. The two sides agree, and this arrow exists for every $1 \leq i \leq m$ .

step 2.2L2

3.4

The arrow $(i ∣ i + 1)$ . Here $t (p) = i + 1$ and $s (p) = i$ , so $p \cdot w_{i} = ((i ∣ i + 1) (i + 1 ∣ i)) \otimes (i ∣ i + 1) = (i ∣ i + 1 ∣ i) \otimes (i ∣ i + 1) = (i ∣ i - 1 ∣ i) \otimes (i ∣ i + 1)$ , using the equality of the two returns at $i$ , legitimate because $0 < i < m$ holds when $1 \leq i \leq m - 1$ , which is exactly the range in which this arrow exists; and $w_{i} \cdot p = ((i) \otimes ((i ∣ i - 1 ∣ i) (i ∣ i + 1))) + ((i ∣ i - 1 ∣ i) \otimes ((i) (i ∣ i + 1))) = (i ∣ i - 1 ∣ i) \otimes (i ∣ i + 1)$ , the first summand being $0$ because $(i ∣ i - 1 ∣ i ∣ i + 1)$ has length three. The two sides agree.

step 2.2L2

3.5

The arrow $(i + 1 ∣ i)$ . Here $t (p) = i$ and $s (p) = i + 1$ , so $p \cdot w_{i} = ((i + 1 ∣ i) (i)) \otimes (i ∣ i - 1 ∣ i) + ((i + 1 ∣ i) (i ∣ i - 1 ∣ i)) \otimes (i) = (i + 1 ∣ i) \otimes (i ∣ i - 1 ∣ i)$ , the second summand vanishing because $(i + 1 ∣ i ∣ i - 1 ∣ i)$ has length three; and $w_{i} \cdot p = ((i + 1 ∣ i) \otimes (i ∣ i + 1)) \cdot p = (i + 1 ∣ i) \otimes ((i ∣ i + 1) (i + 1 ∣ i)) = (i + 1 ∣ i) \otimes (i ∣ i + 1 ∣ i) = (i + 1 ∣ i) \otimes (i ∣ i - 1 ∣ i)$ by the equality of the two returns at $i$ with $0 < i < m$ , again exactly the range $1 \leq i \leq m - 1$ in which this arrow exists. The two sides agree.

step 2.2L2

3.6

The two arrows joining $i - 1$ and $i - 2$ . For $p = (i - 2 ∣ i - 1)$ one has $t (p) = i - 1$ and $s (p) = i - 2$ , so $w_{i} \cdot p = 0$ because no $s_{k}$ equals $i - 2$ , while $p \cdot w_{i} = ((i - 2 ∣ i - 1) (i - 1 ∣ i)) \otimes (i ∣ i - 1) = (i - 2 ∣ i - 1 ∣ i) \otimes (i ∣ i - 1) = 0$ , the monotone path $(i - 2 ∣ i - 1 ∣ i)$ having class $0$ at its interior vertex $i - 1$ , which satisfies $0 < i - 1 < m$ when $i \geq 2$ ; for $p = (i - 1 ∣ i - 2)$ one has $t (p) = i - 2$ and $s (p) = i - 1$ , so $p \cdot w_{i} = 0$ while $w_{i} \cdot p = ((i - 1 ∣ i) \otimes (i ∣ i - 1)) \cdot p = (i - 1 ∣ i) \otimes ((i ∣ i - 1) (i - 1 ∣ i - 2)) = (i - 1 ∣ i) \otimes (i ∣ i - 1 ∣ i - 2) = 0$ , the monotone path $(i ∣ i - 1 ∣ i - 2)$ having class $0$ at its interior vertex $i - 1$ . Both sides agree; both arrows exist only for $i \geq 2$ , and for $i < 2$ this step covers nothing.

step 2.2L2

3.7

The two arrows joining $i + 1$ and $i + 2$ , and the remaining arrows. For $p = (i + 1 ∣ i + 2)$ one has $t (p) = i + 2$ and $s (p) = i + 1$ , so $p \cdot w_{i} = 0$ because no $s_{k}$ equals $i + 2$ , while $w_{i} \cdot p = ((i + 1 ∣ i) \otimes (i ∣ i + 1)) \cdot p = (i + 1 ∣ i) \otimes ((i ∣ i + 1) (i + 1 ∣ i + 2)) = (i + 1 ∣ i) \otimes (i ∣ i + 1 ∣ i + 2) = 0$ , the monotone path $(i ∣ i + 1 ∣ i + 2)$ having class $0$ at its interior vertex $i + 1$ , which satisfies $0 < i + 1 < m$ when $i + 2 \leq m$ ; for $p = (i + 2 ∣ i + 1)$ one has $t (p) = i + 1$ and $s (p) = i + 2$ , so $w_{i} \cdot p = 0$ because no $s_{k}$ equals $i + 2$ , while $p \cdot w_{i} = ((i + 2 ∣ i + 1) (i + 1 ∣ i)) \otimes (i ∣ i + 1) = (i + 2 ∣ i + 1 ∣ i) \otimes (i ∣ i + 1) = 0$ , the monotone path $(i + 2 ∣ i + 1 ∣ i)$ having class $0$ at its interior vertex $i + 1$ . Every remaining arrow $p$ has both endpoints outside ${i - 1, i, i + 1}$ , so neither $t (p)$ nor $s (p)$ equals any $s_{k}$ , and $p \cdot w_{i} = 0 = w_{i} \cdot p$ . All arrows of the quiver are thereby covered.

step 2.2L2

4.1

$γ_{i}$ is a map of $(A_{m}, A_{m})$ -bimodules. By steps 3.1, 3.2, 3.3, 3.4, 3.5, 3.6 and 3.7 the element $w_{i}$ satisfies $a \cdot w_{i} = w_{i} \cdot a$ for every $a \in A_{m}$ , since every element of $A_{m}$ is a finite $Z$ -linear combination of paths and both actions are $Z$ -linear; consequently, for $a, b \in A_{m}$ one has $γ_{i} (b a) = (b a) \cdot w_{i} = b \cdot (a \cdot w_{i}) = b \cdot γ_{i} (a)$ by associativity of the left action, and $γ_{i} (a b) = (a b) \cdot w_{i} = (a \cdot w_{i}) \cdot b = γ_{i} (a) \cdot b$ , the middle equality being the centrality applied to $a \cdot w_{i} \cdot b = a \cdot (w_{i} \cdot b) = a \cdot (b \cdot w_{i})$ . Hence $γ_{i}$ is left and right $A_{m}$ -linear, and by step 1.2 it is degree zero with $γ_{i} (1) = 1 \cdot w_{i} = w_{i}$ of degree $0$ in $U_{i} {- 1}$ .

step 1.2step 3.1step 3.2step 3.3step 3.4step 3.5step 3.6step 3.7

5.1

Natural transformations. Composing $β_{i} \otimes_{A_{m}} {i d}_{M} : U_{i} \otimes_{A_{m}} M \to A_{m} \otimes_{A_{m}} M$ with the unit isomorphism $A_{m} \otimes_{A_{m}} M ≅ M$ of [L5] defines a morphism of $A_{m} -mod$ natural in $M$ , because for a degree-zero map $f : M \to N$ one has $β_{i} (x \otimes y) \cdot f (m) = f (β_{i} (x \otimes y) \cdot m)$ by $A_{m}$ -linearity of $f$ and the unit isomorphisms for $M$ and $N$ are natural by [L5]; thus $β_{i}$ induces a natural transformation $U_{i} (-) \to I d$ , and the identical computation with $w_{i}$ in place of $β_{i}$ , using that $γ_{i}$ is degree zero and left $A_{m}$ -linear by step 4.1, induces a natural transformation $I d \to U_{i} {- 1} (-)$ .

step 1.3step 4.1L5

6.1

Conclusion. The displayed element $w_{i}$ lies in $U_{i}$ and has degree $1$ by steps 1.1 and 1.2, so $γ_{i} (a) = a \cdot w_{i}$ is a well-defined degree-zero map $A_{m} \to U_{i} {- 1}$ with $γ_{i} (1) = w_{i}$ ; it is a map of $(A_{m}, A_{m})$ -bimodules by step 4.1, whose centrality input is the case check of steps 3.1, 3.2, 3.3, 3.4, 3.5, 3.6 and 3.7 over the finitely many arrows, each case using only the equality of the two returns at an interior vertex, the vanishing of monotone length-two paths and the vanishing of all paths of length three; and $β_{i}$ is the degree-zero bimodule map of step 1.3, uniquely determined by $β_{i} (e_{i} \otimes e_{i}) = e_{i}$ by step 2.1. Finally $β_{i}$ and $γ_{i}$ induce the natural transformations of step 5.1, so the bimodule maps $β_{i}$ and $γ_{i}$ of the source's Section 2d are defined, graded of degree zero and bimodule-linear, and the endpoints $i = 1$ and $i = m$ are covered by steps 3.2, 3.3 and 3.5 with the summand of step 1.1 omitted at $i = m$ . All tensor products are over $Z$ or over $A_{m}$ as indicated, the sums involved are finite, and no choice principle is used.

step 1.1step 1.2step 1.3step 2.1step 4.1step 5.1∎

Depends on

Used by

The twist complexes Rᵢ and Rᵢ⁻¹ Definition
Totalizing a two-term twist action Example

Dependency tree · two levels

26 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Mikhail Khovanov and Paul Seidel, Quivers, Floer Cohomology, and Braid Group Actions, §2d, printed pp. 11-12 (standard reference, not scraped)